Differentiate (logx)^x?

1 Answer
Aritra G.
Sep 2, 2017

Let #y = (ln x)^x#

Then taking natural logarithms on both sides,

#ln y = ln (ln x)^x#
#implies ln y = xln (ln x)#

Differentiating both sides,

#1/y(dy)/(dx) = ln (ln x) + x/(ln x)d/(dx)ln x#

#implies (dy)/(dx) = y[ln (ln x) + x/(ln x)1/x]#

#implies (dy)/(dx) = (ln x)^x[ln (ln x) + 1/(ln x)]#

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